Problem 47 - Precalculus Mathematics for Calculus

Question

Find all the real zeros of the polynomial. Use the quadratic formula if necessary, as in Example \(3(a)\). $$P(x)=x^{3}+4 x^{2}+3 x-2$$

Step-by-Step Solution

Answer

The real zeros are \(-2\) and \(-1\) (with multiplicity 2).

Step 1: Finding Possible Rational Roots

First, we apply the Rational Root Theorem which states that any rational root of the polynomial equation \( P(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_0 \) is a fraction \( \frac{p}{q} \), where \( p \) is a factor of the constant term \( a_0 \) and \( q \) is a factor of the leading coefficient \( a_n \). Here, the constant term is \(-2\) and the leading coefficient is \(1\). Thus, the possible rational roots are ±1, ±2.

Step 2: Testing Possible Roots with Synthetic Division

Using synthetic division, test the possible rational roots. We use \( 1 \):\[\begin{array}{r|rrrr}1 & 1 & 4 & 3 & -2 \ & & 1 & 5 & 8 \\hline & 1 & 5 & 8 & 6 \\end{array}\]The remainder is 6, so 1 is not a root. Now test \( -1 \):\[\begin{array}{r|rrrr}-1 & 1 & 4 & 3 & -2 \ & & -1 & -3 & 0 \\hline & 1 & 3 & 0 & -2 \\end{array}\]The remainder is not zero, so \(-1\) is not a root. Repeat this process for \( 2 \) and \( -2 \).

Step 3: Establish 2 as a Root and Solve for Other Roots

When \( 2 \) is tested:\[\begin{array}{r|rrrr}2 & 1 & 4 & 3 & -2 \ & & 2 & 12 & 30 \\hline & 1 & 6 & 15 & 28 \\end{array}\]The remainder is not zero, so \( 2 \) is not a root. Now test \( -2 \):\[\begin{array}{r|rrrr}-2 & 1 & 4 & 3 & -2 \ & & -2 & 0 & -6 \\hline & 1 & 2 & 3 & 0 \\end{array}\]The remainder is 0, so \(-2\) is a root. The quotient polynomial is \( x^2 + 2x + 1 \).

Step 4: Solving the Quotient Polynomial

The quotient is \( x^2 + 2x + 1 \). To find the roots, factor it:\(x^2 + 2x + 1 = (x + 1)^2\).Set \( (x + 1)^2 = 0 \) giving the double root \( x = -1 \).

Step 5: Listing All Real Zeros

The zeros of the polynomial \( P(x) = x^3 + 4x^2 + 3x - 2 \) are found by compiling all different results. By discovering that \(-2\) is a root and \(-1\) is a double root, we establish that the real roots are \(-2, -1, \) and \(-1\) again.

Key Concepts

Understanding the Rational Root Theorem

The Rational Root Theorem is a handy tool when dealing with polynomials. It helps us predict possible rational roots, meaning roots that can be expressed as fractions. This theorem states that if a polynomial \( P(x) = a_nx^n + a_{n-1}x^{n-1} + \, \ldots \, + a_0 \) has a rational root \( x = \frac{p}{q} \), then \( p \) is a factor of the constant term \( a_0 \), and \( q \) is a factor of the leading coefficient \( a_n \).

To use the Rational Root Theorem, list all factors of the constant term and the leading coefficient. For the polynomial \( P(x) = x^3 + 4x^2 + 3x - 2 \):

• Constant term \( a_0 = -2 \), factors: \( \pm 1, \pm 2 \)
• Leading coefficient \( a_n = 1 \), factors: \( \pm 1 \)

Thus, our potential rational roots are \( \pm 1 \) and \( \pm 2 \). These values guide our next steps in testing which, if any, are actual roots of the polynomial.

How to Use Synthetic Division

Synthetic division simplifies the testing of possible roots calculated from the Rational Root Theorem. It's an efficient method for dividing a polynomial by a potential root, especially for linear divisors of the form \( x - c \). In our polynomial case, we can test each potential root by using synthetic division.

Here is how to apply synthetic division:

• Write down the coefficients of the polynomial.
• Select a candidate root.
• Bring down the leading coefficient.
• Multiply the candidate root by the current result and add to the next coefficient.
• Repeat until you reach the final term.

If your last number (the remainder) is zero, the candidate root is indeed a root of the polynomial. For the polynomial \( P(x) = x^3 + 4x^2 + 3x - 2 \), trying \( -2 \) shows it is a root due to a zero remainder.

The synthetic division process not only helps confirm roots but also simplifies the polynomial to a lower degree that can be solved more easily.

Solving with the Quadratic Formula

Once the polynomial is reduced to a quadratic, we can find roots quickly using the Quadratic Formula. Let's refresh how it works. The Quadratic Formula solves equations of the form \( ax^2 + bx + c = 0 \). It is written as:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

This formula yields two solutions, corresponding to the \( \pm \) sign. For the reduced polynomial from prior steps, \( x^2 + 2x + 1 = 0 \), the roots can be found by factoring it as \( (x+1)^2 = 0 \). This reveals a double root, \( x = -1 \). Using the Quadratic Formula confirms this. Plugging in \( a = 1 \), \( b = 2 \), \( c = 1 \), the term under the square root (the discriminant) is zero:\( b^2 - 4ac = 0 \). This zero discriminant confirms a double root solution, aligning with our factorization.Using these methods together helps ensure that you don't miss any potential real roots, and verify them quickly once found.